Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
int main() {
  int r, c;
  cin >> r >> c;
  int i, j;
  int m[105][105];
  int y[105][105];
  memset(y, 0, sizeof(y));
  for (i = 1; i <= r; i++) {
    for (j = 1; j <= c; j++) {
      cin >> m[i][j];
    }
  }
  int ri[105];
  int ra[105];
  int ci[105];
  int ca[105];
  for (i = 1; i <= r; i++) {
    ri[i] = m[i][1];
    ra[i] = m[i][1];
    for (j = 1; j <= c; j++) {
      ri[i] = ((ri[i]) < (m[i][j]) ? (ri[i]) : (m[i][j]));
      ra[i] = ((ra[i]) > (m[i][j]) ? (ra[i]) : (m[i][j]));
    }
  }
  for (i = 1; i <= c; i++) {
    ci[i] = m[1][i];
    ca[i] = m[1][i];
    for (j = 1; j <= r; j++) {
      ci[i] = ((ci[i]) < (m[j][i]) ? (ci[i]) : (m[j][i]));
      ca[i] = ((ca[i]) > (m[j][i]) ? (ca[i]) : (m[j][i]));
    }
  }
  if (r <= c) {
    vector<pair<string, int> > v;
    int ans = -1;
    int cnt = 0;
    int k;
    for (i = 1; i <= r; i++) {
      cnt = cnt + ri[i];
      string s = "row";
      for (k = 1; k <= ri[i]; k++) {
        v.push_back(make_pair(s, i));
      }
    }
    for (i = 1; i <= r; i++) {
      for (j = 1; j <= c; j++) {
        y[i][j] = y[i][j] + ri[i];
      }
    }
    for (j = 1; j <= c; j++) {
      int w = 0;
      for (i = 1; i <= r; i++) {
        w = ((0) > (m[i][j] - y[i][j]) ? (0) : (m[i][j] - y[i][j]));
      }
      cnt = cnt + w;
      string s1 = "col";
      for (k = 1; k <= w; k++) {
        v.push_back(make_pair(s1, j));
      }
      for (i = 1; i <= r; i++) {
        y[i][j] = y[i][j] + w;
      }
    }
    int f = 0;
    for (i = 1; i <= r; i++) {
      for (j = 1; j <= c; j++) {
        if (m[i][j] != y[i][j]) f = 1;
      }
    }
    if (f == 0) {
      ans = cnt;
      cout << ans << "\n";
      for (i = 0; i < v.size(); i++) {
        cout << v[i].first << " " << v[i].second << "\n";
      }
    } else {
      ans = -1;
      cout << ans << "\n";
    }
  } else if (r > c) {
    vector<pair<string, int> > v;
    int ans = -1;
    int cnt = 0;
    int k;
    for (i = 1; i <= c; i++) {
      cnt = cnt + ci[i];
      string s = "col";
      for (k = 1; k <= ci[i]; k++) {
        v.push_back(make_pair(s, i));
      }
    }
    for (i = 1; i <= c; i++) {
      for (j = 1; j <= r; j++) {
        y[j][i] = y[j][i] + ci[i];
      }
    }
    for (j = 1; j <= r; j++) {
      int w = 0;
      for (i = 1; i <= c; i++) {
        w = ((0) > (m[j][i] - y[j][i]) ? (0) : (m[j][i] - y[j][i]));
      }
      cnt = cnt + w;
      string s1 = "row";
      for (k = 1; k <= w; k++) {
        v.push_back(make_pair(s1, j));
      }
      for (i = 1; i <= c; i++) {
        y[j][i] = y[j][i] + w;
      }
    }
    int f = 0;
    for (i = 1; i <= r; i++) {
      for (j = 1; j <= c; j++) {
        if (m[i][j] != y[i][j]) f = 1;
      }
    }
    if (f == 0) {
      ans = cnt;
      cout << ans << "\n";
      for (i = 0; i < v.size(); i++) {
        cout << v[i].first << " " << v[i].second << "\n";
      }
    } else {
      ans = -1;
      cout << ans << "\n";
    }
  }
  return 0;
}